anonymous
  • anonymous
Xlogy-logz.ylogz-logx.zlogx-logy=1 how to prove this??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
it was supposed to x to the power logy-logz y to the power logz-logx z to the power logx-logy
hartnn
  • hartnn
\(\huge x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}\) like this ?
anonymous
  • anonymous
ah thank god..yes like that

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hartnn
  • hartnn
consider \(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}]\) can u simplify that ?
hartnn
  • hartnn
use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)
hartnn
  • hartnn
you should get that =0 = ln 1
anonymous
  • anonymous
i dont understand y u multiplied the whole by log????
hartnn
  • hartnn
to be able to simplify
hartnn
  • hartnn
else, nothing can't be really done to that expression
anonymous
  • anonymous
i have done that problem in this way \[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]
anonymous
  • anonymous
and i cudnt figure out the next step..... :(
hartnn
  • hartnn
i know, i tried that also....since i could not proceed, i thought of taking log....
anonymous
  • anonymous
if u dont mind,can u show me the simplification process...??? :)
hartnn
  • hartnn
\(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}] \\\large =\log[x^{\log y-\log z}]+\log[y^{\log z-\log x}]+\log[z^{\log x-\log y}]\) got this ?
hartnn
  • hartnn
now use \( \\ \log A^{B}=B\log A\)
hartnn
  • hartnn
@suss
anonymous
  • anonymous
oh sorry man i jus lost connection leeme check it out
anonymous
  • anonymous
y did u keep the + sign ???
anonymous
  • anonymous
@hartnn
hartnn
  • hartnn
because \(\log AB=\log A+\log B \\ \)
anonymous
  • anonymous
ah finally got it .....thx a ton man
hartnn
  • hartnn
sure ? u getting that = 0 ? then write 0=log 1
anonymous
  • anonymous
yes 100%..i hav to use formula of log ab bak ther ......i really got it man
hartnn
  • hartnn
good :) welcome ^_^

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